Homework 7, Mathematics 1
submit by 28.11.
Only problems 2cef 3, 4 and 6b will be graded.
Problem 1:
(a) Prove that
(arccos x)0 = − √
1
1 − x2
for x ∈ (−1, 1).
(b)* Starting from the definition of the derivative, prove that
(arccos x)0+ −1 = −∞,
(arccos x)0− = −∞.
1
(arccos x)0+ −1 ,
x−π
Hint: For
the definition leads to computing limx→−1+ arccos
.
x+1
Use substitution y = arccos x − π and pay attention to from what side y approaches 0 (you can use the corresponding substitution theorem, even if we did
not formulate it).
(c) Prove that
1
(arccotg x)0 = −
for x ∈ R.
1 + x2
Solution:
(a) Let arccos x = y, y ∈ (0, π). By the theorem for the derivative of the inverse,
since (cos y)0 = − sin y 6= 0 for x ∈ (0, π), we get
(arccos x)0 =
1
1
1
1
= −p
= −√
=−
.
(cos y)0
sin y
1 − x2
1 − (cos y)2
(b) As in the hint, with y = arccos x − π → 0− as x → −1+ ,
arccos x − arccos(−1)
x − (−1)
arccos x − π
= lim
x→−1+
x+1
y
= lim
y→0− cos(y + π) + 1
y
= lim
y→0− − cos y + 1
1
y2
·
= lim
y→0− 1 − cos y y
= −∞.
(arccos x)0+ −1 =
lim
x→−1+
1
In a similar way, with y = arccos x → 0+ as x → 1− ,
arccos x − arccos(1)
(arccos x)0− 1 = lim
x→1−
x−1
arccos x
= lim
x→1− x − 1
y
= lim
y→0+ cos y − 1
y2
1
= lim −
·
y→0+
1 − cos y y
= −∞.
(c) Let arccotg x = y, y ∈ (0, π). By the theorem for the derivative of the
inverse, since (cotg y)0 = − sin12 x 6= 0 for x ∈ (0, π), we get
(arccotg x)0 =
1
1
1
1
1
= − 1 = − sin2 y+cos2 y = −
2 = − 1 + x2 .
(cotg y)0
1
+
cotg
y
2
sin2 y
sin y
Problem 2: Find the definition domain of the following functions and compute
their derivative at the points where it exists:
(a)
f (x) = ln(1 + cos x),
1
f (x) = arccos
,
x
(b)
(c) [3 points]
(e) [3 points]
f (x) = arctan(ln x),
x cos(3x)
f (x) = ln tan
−
,
2
2 sin2 x
f (x) = xx ,
(f ) [4 points]
f (x) = (sin x)cos x + (cos x)sin x .
(d)
Hint: Express any f (x)g(x) as eln(f (x))·g(x) .
Solution:
(a) Definition domain: 1 + cos x > 0 ⇔ cos x 6= −1 ⇔ x 6= 2kπ + 23 π, k ∈ Z
For x ∈ R \ {2kπ + 32 π : k ∈ Z},
f 0 (x) =
(b) Definition domain:
(−∞, −1) ∪ (1, ∞),
f 0 (x) = − q
1
1−
1
x
1
sin x
· (cos x)0 = −
.
1 + cos x
1 + cos x
∈ [−1, 1] ⇔ x ∈ (−∞, −1] ∪ [1, ∞).
−1
· x
1 2
0
= −√
x
2
For x ∈
|x|
1
√
· (−1) · x−2 =
.
2
x −1
|x| x2 − 1
For x = −1, after substitution y =
0
f−
(−1) =
lim
x→−1−
= lim
y→−1+
1
x
(x 6= −1 =⇒ y 6= −1!), we get
arccos y − π
arccos(1/x) − π
= lim
1
y→−1+
x+1
y +1
arccos y − π
arccos y − arccos(−1)
· y = lim
· y = +∞,
y→−1
1+y
y − (−1)
+
since limy→−1+ arccos(y)−arccos(−1)
= (arccos y)0+ (−1) = −∞ and limy→−1+ y =
y−(−1)
0
−1. Similarly, f (1)+ = +∞.
(c) Definition domain: x > 0. For such x,
f 0 (x) =
1
1
· (ln x)0 =
.
2
1 + ln x
x(1 + ln2 x)
(d) Definition domain: tan x2 > 0 and sin x 6= 0 ⇔ x ∈ ∪k∈Z (kπ, kπ + π2 ). Then
x 0 − sin(3x) · (3x)0 · (2 sin2 x) − cos(3x) · 2 · 2 sin x · (sin x)0
1
·
tan
−
tan x2
2
(2 sin2 x)2
x 0 sin(3x) · 3 · (2 sin2 x) + cos(3x) · 2 · 2 sin x · cos x
1
1
+
=
·
x ·
2
tan 2 1 + x
2
4 sin4 x
f 0 (x) =
2
3 sin(3x) sin x + 4 cos(3x) cos x
2x
+
=
(4 + x2 ) tan x2
2 sin3 x
=
2x
3
3 cos2 x
cos4 x
−
sin
x
−
+
2
.
x
(4 + x2 ) tan 2
2
2 sin x
sin3 x
(e) Definition domain: The general exponential function y α , if α ∈
/ Z, is defined
for y > 0. Hence xx makes sense for either x ∈ Z, x 6= 0 (optionally, 00 is
sometimes defined as well, as 00 = 1), or x ∈ R. (Note: {x : x > 0} will also be
taken as right answer.) In order to differentiate, the function mmust be defined
in a neighborhood around the point. Hence, it only makes sence to speak of f 0
only for x > 0. Then
f 0 (x) = (ex·ln x )0 = ex ln x · (x ln x)0 = xx (1 · ln x + x ·
1
) = xx (ln x + 1).
x
(f ) Definition domain: sin x > 0 and cos x > 0 , or poosible when both sin x ∈ Z
and cos x ∈ Z. The second option happens additionally only when sin x ∈
{0, −1} or cos x ∈ {0, −1}; however, if sin x = −1 or cos x = −1, then cos x = 0
or sin x = 0, and 0−1 is not defined. Hence, the definition domain is
D = {x ∈ R : sin x > 0, cos x > 0}∪
{x : sin x = 0 and cos x = 1, or sin x = 1 and cos x = 0}
[
π
=
[2kπ, 2kπ + ].
2
k∈Z
3
For x ∈
S
k∈Z (2kπ, 2kπ
+ π2 ), we have
f 0 (x) =(ecos x·ln(sin x) + esin x ln(cos x) )0
=ecos x·ln(sin x) (cos x · ln(sin x)0 + esin x ln(cos x) (sin x ln(cos x))0
1
cos x
0
=(sin x)
− sin x ln(sin x) + cos x
(sin x)
sin x
1
sin x
0
+ (cos x)
cos x ln(cos x) + sin x
(cos x)
cos x
cos2 x
=(sin x)cos x − sin x ln(sin x) +
sin x
sin2 x
.
+ (cos x)sin x cos x ln(cos x) −
cos x
0
0
(2kπ + π2 ) is more involved, one gets 0(= 1 − 1) as
(2kπ) and f−
Computing f+
a result.
√
Problem 3 [2×3 points]: Let f (x) = ln |x + x2 + a|. Find the definition
domain of f and the derivative f 0 at points where it exists (no need to compute
one-sided derivatives at endpoints), if
(a) a > 0;
(b) a ≤ 0.
Solution:
(a) Definition domain R, f 0 (x) = √x12 +a .
(b) a = 0: Definition domain x 6= 0. √
√
a < 0: Definition domain (−∞,
√ − −a]
√∪ [ a, ∞).
In any case, for x ∈ (−∞, − −a) ∪ ( a, ∞): f 0 (x) = √x12 +a (in computa√
√
tion, one needs to distinguish the cases x + x2 + a > 0 and x + x2 + a < 0).
Problem 4 [5 points]: Compute the first and the second derivatives of the
following function for each x ∈ R. Does the third derivative exist at x = 0?
Justify your answer.
(
x4 sin x1
if x 6= 0,
f (x) =
0
if x = 0.
Solution:
x 6= 0 =⇒ f 0 (x) = 4x3 sin
1
1
+ x4 cos
x
x
0
1
1
1
= 4x3 sin − x2 cos ,
x
x
x
x4 sin x1 − 0
1
= lim x3 sin = 0
x→0
x→0
x−0
x
f 0 (0) = lim
4
by the sandwich theorem (as −|x|3 ≤ x3 sin x1 ≤ |x|3 ). Then, in a similar way,
0
0
1
1 1
1
1 1
00
2
3
2
x 6= 0 =⇒ f (x) = 12x sin + 4x cos
− 2x cos + x sin
x
x x
x
x x
1
1
1
= 12x2 sin − 6x cos − sin ,
x
x
x
0
0
f
(x)
−
f
(0)
1
1
f 0 (0) = lim
= lim 4x2 sin − x cos = 0
x→0
x→0
x−0
x
x
again by the sandwich theorem, as −4x2 − |x| ≤ 4x2 sin x1 − x cos x1 ≤ 4x2 + |x|.
Notice that f 00 is not continuous at 0, so f is not differentiable at 0.
Problem 5: Compute (x2 cos x)(47) .
Solution: By induction,
(cos x)(n)

cos x



− sin x
=

− cos x



sin x
if
if
if
if
n = 4k, k ∈ N0
n = 4k + 1, k ∈ N0
n = 4k + 2, k ∈ N0
n = 4k + 3, k ∈ N0 .
By the Leibniz formula, as (x2 )(n) = 0 for any n ≥ 3, we have
47 2
47
47
2
(47)
(47)
2 0
(46)
x (cos x)
=
x (cos x)
+
(x ) (cos x)
+
(x2 )00 (cos x)(45)
0
1
2
= x2 sin x − 96x cos x − 47 · 46 sin x.
Problem 6: Find general formula for the following derivatives for all n ∈ N.
Express with no derivatives and simplify as much as possible.
(xn ex )(n) ;
(n)
ln x
.
x
(a)
(b) [3* bonus points]
Solution: (a) Using the Leibniz formula,
n X
n
(xn ex )(n) =
(xn )(k) (ex )(n−k)
k
k=0
n X
n
=
n · (n − 1) · . . . · (n − k + 1)xn−k ex
k
=
=
k=0
n
X
k=0
n
X
k=0
(n!)2
xn−k ex
(n − k)!(k!)2
(n!)2
xk ex .
((n − k)!)2 k!
5
(b) Byinduction, for any n ∈ N,
(x−1 )(n) = (−1)n n!x−n−1
and
(ln x)(n) = (−1)n+1 (n − 1)!x−n .
Hence, by Leibniz formula,
ln x
x
(n)
=
=
n X
n
k=0
n X
k=0
k
(ln x)(k) (x−1 )(n−k)
n
(−1)k+1 (k − 1)!x−k (−1)n−k (n − k)!x−n+k−1
k
n
=
(−1)n+1 n! X 1
.
xn+1
k
k=0
6